Solve for $x$ : $ 3|x - 3| - 3 = 1|x - 3| + 6 $
Solution: Subtract $ {1|x - 3|} $ from both sides: $ \begin{eqnarray} 3|x - 3| - 3 &=& 1|x - 3| + 6 \\ \\ { - 1|x - 3|} && { - 1|x - 3|} \\ \\ 2|x - 3| - 3 &=& 6 \end{eqnarray} $ Add ${3}$ to both sides: $ \begin{eqnarray} 2|x - 3| - 3 &=& 6 \\ \\ { + 3} &=& { + 3} \\ \\ 2|x - 3| &=& 9 \end{eqnarray} $ Divide both sides by ${2}$ $ \dfrac{2|x - 3|} {{2}} = \dfrac{9} {{2}} $ Simplify: $ |x - 3| = \dfrac{9}{2}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 3 = -\dfrac{9}{2} $ or $ x - 3 = \dfrac{9}{2} $ Solve for the solution where $x - 3$ is negative: $ x - 3 = -\dfrac{9}{2} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& -\dfrac{9}{2} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& -\dfrac{9}{2} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $2$ $ x = - \dfrac{9}{2} {+ \dfrac{6}{2}} $ $ x = -\dfrac{3}{2} $ Then calculate the solution where $x - 3$ is positive: $ x - 3 = \dfrac{9}{2} $ Add ${3}$ to both sides: $ \begin{eqnarray} x - 3 &=& \dfrac{9}{2} \\ \\ {+ 3} && {+ 3} \\ \\ x &=& \dfrac{9}{2} + 3 \end{eqnarray} $ Change the ${ + 3}$ to an equivalent fraction with a denominator of $2$ $ x = \dfrac{9}{2} {+ \dfrac{6}{2}} $ $ x = \dfrac{15}{2} $ Thus, the correct answer is $x = -\dfrac{3}{2} $ or $x = \dfrac{15}{2} $.